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Question-200533

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Question Number 200533 by ajfour last updated on 19/Nov/23
Commented by mr W last updated on 20/Nov/23
AB⊥AD is given? otherwise there is no unique solution.
$${AB}\bot{AD}\:{is}\:{given}? \\ $$$${otherwise}\:{there}\:{is}\:{no}\:{unique}\:{solution}. \\ $$
Commented by mr W last updated on 20/Nov/23
Commented by ajfour last updated on 20/Nov/23
p+q=13 wont do sir, obvious!
$${p}+{q}=\mathrm{13}\:\:\:{wont}\:{do}\:{sir},\:{obvious}! \\ $$
Commented by ajfour last updated on 20/Nov/23
yes sir given AB⊥AD. forgot to mark so.
$${yes}\:{sir}\:{given}\:{AB}\bot{AD}.\:{forgot}\:{to} \\ $$$${mark}\:{so}. \\ $$
Commented by a.lgnaoui last updated on 20/Nov/23
look at the exact answer ↓
$$\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{exact}\:\:\mathrm{answer}\:\:\downarrow \\ $$
Answered by a.lgnaoui last updated on 20/Nov/23
NB=15−7=8 DE=MD=12−7=5 CE=p−5 CN=q−8 q^2 =OB^2 +OC^2 −2OB×OCcos 𝛌 (1) 𝛌=∡(BOC)=∡BON+∡NOC=𝛂+𝛃 tan (∡BON)=(8/7) tan(∡ NOC)=((q−8)/7) ⇒(1/(cos^2 α))=1+((64)/(49))=((113)/(49)) cos 𝛂=((7(√(113)))/(113)) (1/(cos^2 𝛃))=1+(((q−8)^2 )/7^2 )=((49+(q−8)^2 )/(49)) cos β=(7/( (√(49+(q−8)^2 )))) CE=CN ⇒p−5=q−8 ⇒ q−p=3 (2) (1) q^2 =[(49+64)]+[(49+(q−8)^2 ] 2[((√(113)) ×(√(49+q−8)^2 )) ]cos 𝛌 cos 𝛌=cos α𝛂cos 𝛃−sin 𝛂αsin 𝛃 { ((sin 𝛂=((8(√(113)))/( 113)))),((sin 𝛃 =((q−8)/( (√(49+(q−8)^2 )))))) :} ⇒q^2 =(113+49)+(q−8)^2 −2×(√(113))(((49−8(q−8))/( 1]))) q^2 =162+(q−8)^2 )−2(√(113)) [49−8(q−8)] q^2 =(q−8)^2 −16(√(113 )) (q−8) +98(√(113)) −162 posons x=q−8 (x+8)^2 −x^2 −16x(√(113)) +879,754=0 16x−16x(√(113)) +879,754=0 16x((√(113)) −1)=879,754 x=((879,754)/( 154))=5,712 ⇒q=8+5,712=13,712 (2)⇒p= q−3 ⇒ p=10,712 alors p+q=24,424
$$\:\boldsymbol{\mathrm{NB}}=\mathrm{15}−\mathrm{7}=\mathrm{8}\:\:\:\:\:\boldsymbol{\mathrm{DE}}=\boldsymbol{\mathrm{MD}}=\mathrm{12}−\mathrm{7}=\mathrm{5} \\ $$$$\boldsymbol{\mathrm{CE}}=\boldsymbol{\mathrm{p}}−\mathrm{5}\:\:\:\:\:\:\:\boldsymbol{\mathrm{CN}}=\boldsymbol{\mathrm{q}}−\mathrm{8} \\ $$$$\boldsymbol{\mathrm{q}}^{\mathrm{2}} =\boldsymbol{\mathrm{OB}}^{\mathrm{2}} +\boldsymbol{\mathrm{OC}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{OB}}×\boldsymbol{\mathrm{OC}}\mathrm{cos}\:\boldsymbol{\lambda}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\:\:\:\boldsymbol{\lambda}=\measuredangle\left(\boldsymbol{\mathrm{BOC}}\right)=\measuredangle\boldsymbol{\mathrm{BON}}+\measuredangle\boldsymbol{\mathrm{NOC}}=\boldsymbol{\alpha}+\boldsymbol{\beta} \\ $$$$\:\:\mathrm{tan}\:\left(\measuredangle\boldsymbol{\mathrm{BON}}\right)=\frac{\mathrm{8}}{\mathrm{7}}\:\:\:\:\mathrm{tan}\left(\measuredangle\:\boldsymbol{\mathrm{NOC}}\right)=\frac{\boldsymbol{\mathrm{q}}−\mathrm{8}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \alpha}=\mathrm{1}+\frac{\mathrm{64}}{\mathrm{49}}=\frac{\mathrm{113}}{\mathrm{49}}\:\:\:\:\mathrm{cos}\:\boldsymbol{\alpha}=\frac{\mathrm{7}\sqrt{\mathrm{113}}}{\mathrm{113}} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\beta}}=\mathrm{1}+\frac{\left(\mathrm{q}−\mathrm{8}\right)^{\mathrm{2}} }{\mathrm{7}^{\mathrm{2}} }=\frac{\mathrm{49}+\left(\mathrm{q}−\mathrm{8}\right)^{\mathrm{2}} }{\mathrm{49}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\beta=\frac{\mathrm{7}}{\:\sqrt{\mathrm{49}+\left(\mathrm{q}−\mathrm{8}\right)^{\mathrm{2}} }} \\ $$$$\boldsymbol{\mathrm{CE}}=\boldsymbol{\mathrm{CN}} \\ $$$$\:\Rightarrow\mathrm{p}−\mathrm{5}=\mathrm{q}−\mathrm{8}\:\:\:\:\Rightarrow\:\boldsymbol{\mathrm{q}}−\boldsymbol{\mathrm{p}}=\mathrm{3}\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\:\boldsymbol{\mathrm{q}}^{\mathrm{2}} =\left[\left(\mathrm{49}+\mathrm{64}\right)\right]+\left[\left(\mathrm{49}+\left(\boldsymbol{\mathrm{q}}−\mathrm{8}\right)^{\mathrm{2}} \right]\right. \\ $$$$\:\:\:\:\:\:\:\mathrm{2}\left[\left(\sqrt{\mathrm{113}}\:×\sqrt{\left.\mathrm{49}+\boldsymbol{\mathrm{q}}−\mathrm{8}\right)^{\mathrm{2}} }\:\right]\mathrm{cos}\:\boldsymbol{\lambda}\right. \\ $$$$\mathrm{cos}\:\boldsymbol{\lambda}=\mathrm{cos}\:\alpha\boldsymbol{\alpha}\mathrm{cos}\:\boldsymbol{\beta}−\mathrm{sin}\:\boldsymbol{\alpha}\alpha\mathrm{sin}\:\boldsymbol{\beta} \\ $$$$\begin{cases}{\mathrm{sin}\:\boldsymbol{\alpha}=\frac{\mathrm{8}\sqrt{\mathrm{113}}}{\:\mathrm{113}}}\\{\mathrm{sin}\:\boldsymbol{\beta}\:=\frac{\mathrm{q}−\mathrm{8}}{\:\sqrt{\mathrm{49}+\left(\mathrm{q}−\mathrm{8}\right)^{\mathrm{2}} }}}\end{cases} \\ $$$$\Rightarrow\boldsymbol{\mathrm{q}}^{\mathrm{2}} =\left(\mathrm{113}+\mathrm{49}\right)+\left(\boldsymbol{\mathrm{q}}−\mathrm{8}\right)^{\mathrm{2}} \\ $$$$\:−\mathrm{2}×\sqrt{\mathrm{113}}\left(\frac{\mathrm{49}−\mathrm{8}\left(\boldsymbol{\mathrm{q}}−\mathrm{8}\right)}{\left.\:\mathrm{1}\right]}\right) \\ $$$$ \\ $$$$\left.\boldsymbol{\mathrm{q}}^{\mathrm{2}} \:=\mathrm{162}+\left(\boldsymbol{\mathrm{q}}−\mathrm{8}\right)^{\mathrm{2}} \right)−\mathrm{2}\sqrt{\mathrm{113}}\:\left[\mathrm{49}−\mathrm{8}\left(\boldsymbol{\mathrm{q}}−\mathrm{8}\right)\right] \\ $$$$ \\ $$$$\boldsymbol{\mathrm{q}}^{\mathrm{2}} =\left(\boldsymbol{\mathrm{q}}−\mathrm{8}\right)^{\mathrm{2}} −\mathrm{16}\sqrt{\mathrm{113}\:}\:\left(\boldsymbol{\mathrm{q}}−\mathrm{8}\right) \\ $$$$\:\:+\mathrm{98}\sqrt{\mathrm{113}}\:\:−\mathrm{162} \\ $$$$\boldsymbol{\mathrm{posons}}\:\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{q}}−\mathrm{8} \\ $$$$ \\ $$$$\left(\boldsymbol{\mathrm{x}}+\mathrm{8}\right)^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{16}\boldsymbol{\mathrm{x}}\sqrt{\mathrm{113}}\:+\mathrm{879},\mathrm{754}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{16}\boldsymbol{\mathrm{x}}−\mathrm{16x}\sqrt{\mathrm{113}}\:+\mathrm{879},\mathrm{754}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{16}\boldsymbol{\mathrm{x}}\left(\sqrt{\mathrm{113}}\:−\mathrm{1}\right)=\mathrm{879},\mathrm{754} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{879},\mathrm{754}}{\:\mathrm{154}}=\mathrm{5},\mathrm{712} \\ $$$$\Rightarrow\boldsymbol{\mathrm{q}}=\mathrm{8}+\mathrm{5},\mathrm{712}=\mathrm{13},\mathrm{712} \\ $$$$ \\ $$$$\:\:\:\:\:\left(\mathrm{2}\right)\Rightarrow\boldsymbol{\mathrm{p}}=\:\:\boldsymbol{\mathrm{q}}−\mathrm{3} \\ $$$$\:\:\:\:\:\Rightarrow\:\boldsymbol{\mathrm{p}}=\mathrm{10},\mathrm{712} \\ $$$$ \\ $$$$\mathrm{alors}\:\:\:\:\:\:\:\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}=\mathrm{24},\mathrm{424} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 20/Nov/23
Commented by ajfour last updated on 20/Nov/23
I ll check your solution in detail, there must be some mistake, perhaps. huge values!
Answered by mr W last updated on 20/Nov/23
Commented by mr W last updated on 20/Nov/23
7(tan α+tan β)=15 7(tan α+tan γ)=12 7(tan γ+tan δ)=p 7(tan β+tan δ)=q p+q=7(tan β+tan γ+2 tan δ) δ=π−(α+β+γ) tan δ=−((tan (α+β)+tan γ)/(1−tan (α+β)tan γ)) =−((((tan α+tan β)/(1−tan α tan β))+tan γ)/(1−((tan α+tan β)/(1−tan α tan β))×tan γ)) =((tan α+tan β+tan γ−tan α tan β tan γ)/(tan α tan β+tan β tan γ+tan γ tan α−1)) given: α=45°, tan α=1 tan β=((15)/7)−tan α=(8/7) tan γ=((12)/7)−tan α=(5/7) tan δ=((1+(8/7)+(5/7)−1×(8/7)×(5/7))/(1×(8/7)+(8/7)×(5/7)+(5/7)×1−1))=((50)/(41)) p=7×((5/7)+((50)/(41)))=((555)/(41)) q=7×((8/7)+((50)/(41)))=((678)/(41)) p+q=((1233)/(41))≈30.07
$$\mathrm{7}\left(\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta\right)=\mathrm{15} \\ $$$$\mathrm{7}\left(\mathrm{tan}\:\alpha+\mathrm{tan}\:\gamma\right)=\mathrm{12} \\ $$$$\mathrm{7}\left(\mathrm{tan}\:\gamma+\mathrm{tan}\:\delta\right)={p} \\ $$$$\mathrm{7}\left(\mathrm{tan}\:\beta+\mathrm{tan}\:\delta\right)={q} \\ $$$${p}+{q}=\mathrm{7}\left(\mathrm{tan}\:\beta+\mathrm{tan}\:\gamma+\mathrm{2}\:\mathrm{tan}\:\delta\right) \\ $$$$ \\ $$$$\delta=\pi−\left(\alpha+\beta+\gamma\right) \\ $$$$\mathrm{tan}\:\delta=−\frac{\mathrm{tan}\:\left(\alpha+\beta\right)+\mathrm{tan}\:\gamma}{\mathrm{1}−\mathrm{tan}\:\left(\alpha+\beta\right)\mathrm{tan}\:\gamma} \\ $$$$\:\:=−\frac{\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}+\mathrm{tan}\:\gamma}{\mathrm{1}−\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}×\mathrm{tan}\:\gamma} \\ $$$$\:\:=\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta+\mathrm{tan}\:\gamma−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma}{\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta+\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma+\mathrm{tan}\:\gamma\:\mathrm{tan}\:\alpha−\mathrm{1}} \\ $$$${given}:\:\alpha=\mathrm{45}°,\:\mathrm{tan}\:\alpha=\mathrm{1} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{15}}{\mathrm{7}}−\mathrm{tan}\:\alpha=\frac{\mathrm{8}}{\mathrm{7}} \\ $$$$\mathrm{tan}\:\gamma=\frac{\mathrm{12}}{\mathrm{7}}−\mathrm{tan}\:\alpha=\frac{\mathrm{5}}{\mathrm{7}} \\ $$$$\mathrm{tan}\:\delta=\frac{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{7}}+\frac{\mathrm{5}}{\mathrm{7}}−\mathrm{1}×\frac{\mathrm{8}}{\mathrm{7}}×\frac{\mathrm{5}}{\mathrm{7}}}{\mathrm{1}×\frac{\mathrm{8}}{\mathrm{7}}+\frac{\mathrm{8}}{\mathrm{7}}×\frac{\mathrm{5}}{\mathrm{7}}+\frac{\mathrm{5}}{\mathrm{7}}×\mathrm{1}−\mathrm{1}}=\frac{\mathrm{50}}{\mathrm{41}} \\ $$$${p}=\mathrm{7}×\left(\frac{\mathrm{5}}{\mathrm{7}}+\frac{\mathrm{50}}{\mathrm{41}}\right)=\frac{\mathrm{555}}{\mathrm{41}} \\ $$$${q}=\mathrm{7}×\left(\frac{\mathrm{8}}{\mathrm{7}}+\frac{\mathrm{50}}{\mathrm{41}}\right)=\frac{\mathrm{678}}{\mathrm{41}} \\ $$$${p}+{q}=\frac{\mathrm{1233}}{\mathrm{41}}\approx\mathrm{30}.\mathrm{07} \\ $$
Commented by mr W last updated on 20/Nov/23
Commented by ajfour last updated on 20/Nov/23
Thanks Sir, utter good presented!
$${Thanks}\:{Sir},\:{utter}\:{good}\:{presented}! \\ $$
Commented by justenspi last updated on 21/Nov/23
Sir how can I contact you , I am available through any social media app. Thanks in advance.
$$\mathrm{Sir}\:\mathrm{how}\:\mathrm{can}\:\mathrm{I}\:\mathrm{contact}\:\mathrm{you}\:,\:\mathrm{I}\:\mathrm{am}\:\mathrm{available} \\ $$$$\mathrm{through}\:\mathrm{any}\:\mathrm{social}\:\mathrm{media}\:\mathrm{app}.\:\mathrm{Thanks}\:\mathrm{in}\:\mathrm{advance}. \\ $$
Commented by mr W last updated on 21/Nov/23
but i′m available only in this forum. when you think i can help you, then post your questions here. i′ll try my best.
$${but}\:{i}'{m}\:{available}\:{only}\:{in}\:{this}\:{forum}. \\ $$$${when}\:{you}\:{think}\:{i}\:{can}\:{help}\:{you},\:{then} \\ $$$${post}\:{your}\:{questions}\:{here}.\:{i}'{ll}\:{try}\:{my} \\ $$$${best}. \\ $$
Commented by justenspi last updated on 21/Nov/23
Thanks sir really appreciate ur time I am in middle school , I wanted to ask you how to get better at trigonometry and geometry in that manner , are there any books or problem books I should do !
$${Thanks}\:{sir}\:{really}\:{appreciate}\:{ur}\:{time} \\ $$$$\:{I}\:{am}\:{in}\:{middle}\:{school}\:,\:{I}\:{wanted}\:{to} \\ $$$${ask}\:{you}\:{how}\:{to}\:{get}\:{better}\:{at}\:{trigonometry} \\ $$$${and}\:{geometry}\:{in}\:{that}\:{manner}\:,\:{are}\:{there}\:{any} \\ $$$${books}\:{or}\:{problem}\:{books}\:{I}\:{should}\:{do}\:! \\ $$
Commented by mr W last updated on 21/Nov/23
i′ve left the school for a long time, i really don′t know what books i can recommend.
$${i}'{ve}\:{left}\:{the}\:{school}\:{for}\:{a}\:{long}\:{time}, \\ $$$${i}\:{really}\:{don}'{t}\:{know}\:{what}\:{books}\:{i}\:{can} \\ $$$${recommend}. \\ $$
Commented by justenspi last updated on 21/Nov/23
Okay sir , thanks 😊
$${Okay}\:{sir}\:,\:{thanks}\: \\ $$😊

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